An electron in a singly ionized helium atom ( He^+ ) is in an...

Given, initial angular momentum $L_i = \dfrac{2h}{\pi} = n_i \dfrac{h}{2\pi} \Rightarrow n_i = 4$

Final angular momentum $L_f = \dfrac{h}{\pi} = n_f \dfrac{h}{2\pi} \Rightarrow n_f = 2$

For a singly ionized helium atom ($He^+$), the atomic number $Z = 2$.

The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:

$\dfrac{1}{\lambda} = R Z^2 \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)$

$\Rightarrow \dfrac{1}{\lambda} = R (2)^2 \left( \dfrac{1}{2^2} - \dfrac{1}{4^2} \right)$

$\Rightarrow \dfrac{1}{\lambda} = 4R \left( \dfrac{1}{4} - \dfrac{1}{16} \right)$

$\Rightarrow \dfrac{1}{\lambda} = 4R \left( \dfrac{3}{16} \right) = \dfrac{3R}{4}$

$\Rightarrow \lambda = \dfrac{4}{3R}$

Using $Z=1$ incorrectly leads to $\dfrac{16}{3R}$, and misinterpreting the angular momentum values as $n=2$ and $n=1$ leads to $\dfrac{1}{3R}$.

Answer: $\dfrac{4}{3R}$